USACO月赛—JAN12 Silver

题目描述
After several years of record milk production, Farmer John now operates an entire network of N farms (1 <= N <= 100).  Farm i is located at position (x_i, y_i) in the 2D plane, distinct from all other farms, with both x_i and y_i being integers.

FJ needs your help planning his daily delivery route to bring supplies to the N farms.  Starting from farm 1, he plans to visit the farms sequentially (farm 1, then farm 2, then farm 3, etc.), finally returning to farm 1 after visiting farm N.  It tames FJ one minute to make a single step either north, south, east, or west.  Furthermore, FJ wants to visit each farm exactly once during his entire journey (except farm 1, which he visits twice of course).  

Please help FJ determine the minimum amount of time it will take him to complete his entire delivery route.

输入
* Line 1: The number of farms, N.

* Lines 2..1+N: Line i+1 contains two space-separated integers, x_i and y_i  (1 <= x_i, y_i <= 1,000,000).
输出
* Line 1: The minimum number of minutes FJ needs to complete his delivery route, or -1 if it is impossible to find a feasible delivery route that visits every farm exactly once (except farm 1).

样例输入
4
2 2
2 4
2 1
1 3
样例输出
12
提示
OUTPUT DETAILS:



FJ can complete his delivery route in 12 minutes: 2 minutes to go from farm 1 to farm 2, 5 minutes to go from farm 2 to farm 3 (circumventing farm 1), 3 minutes to go from farm 3 to farm 4, and then 2 minutes to return to farm 1.

题目描述
Farmer John has just received a new shipment of N (1 <= N <= 20) bales of hay, where bale i has size S_i (1 <= S_i <= 100).  He wants to divide the bales between his three barns as fairly as possible.  

After some careful thought, FJ decides that a "fair" division of the hay bales should make the largest share as small as possible.  That is, if B_1, B_2, and B_3 are the total sizes of all the bales placed in barns 1, 2, and 3, respectively (where B_1 >= B_2 >= B_3), then FJ wants to make B_1 as small as possible.

For example, if there are 8 bales in these sizes:

2 4 5 8 9 14 15 20

A fair solution is

Barn 1: 2 9 15   B_1 = 26
Barn 2: 4 8 14   B_2 = 26
Barn 3: 5 20     B_3 = 25

Please help FJ determine the value of B_1 for a fair division of the hay bales.


输入
* Line 1: The number of bales, N.

* Lines 2..1+N: Line i+1 contains S_i, the size of the ith bale.

输出
* Line 1: Please output the value of B_1 in a fair division of the hay  bales.

样例输入
8
14
2
5
15
8
9
20
4
样例输出
26

题目描述
Farmer John has discovered that his cows produce higher quality milk when they are subject to strenuous exercise.  He therefore decides to send his N cows (1 <= N <= 25,000) to climb up and then back down a nearby mountain!

Cow i takes U(i) time to climb up the mountain and then D(i) time to climb down the mountain.  Being domesticated cows, each cow needs the help of a farmer for each leg of the climb, but due to the poor economy, there are only two farmers available, Farmer John and his cousin Farmer Don.  FJ plans to guide cows for the upward climb, and FD will then guide the cows for the downward climb.  Since every cow needs a guide, and there is only one farmer for each part of the voyage, at most one cow may be climbing upward at any point in time (assisted by FJ), and at most one cow may be climbing down at any point in time (assisted by FD).  A group of cows may temporarily accumulate at the top of the mountain if they climb up and then need to wait for FD‘s assistance before climbing down.  Cows may climb down in a different order than they climbed up.

Please determine the least possible amount of time for all N cows to make the entire journey.

输入
* Line 1: The number of cows, N.

* Lines 2..1+N: Line i+1 contains two space-separated integers: U(i) and D(i).  (1 <= U(i), D(i) <= 50,000).

输出
* Line 1: A single integer representing the least amount of time for all the cows to cross the mountain.
样例输入
3
6 4
8 1
2 3
样例输出
17
提示
OUTPUT DETAILS:



If cow 3 goes first, then cow 1, and then cow 2 (and this same order is used for both the ascent and descent), this gives a total time of 17.

  1 #include<set>
  2 #include<queue>
  3 #include<vector>
  4 #include<cstdio>
  5 #include<cstring>
  6 #include<cstdlib>
  7 #include<iostream>
  8 #include<algorithm>
  9 const int N = 110;
 10 using namespace std;
 11 const int limits = 1000000;
 12 typedef pair<int,int> Points;
 13 const int dx[5] = {0,0,1,-1,0},
 14           dy[5] = {0,1,0,0,-1};
 15 #define x first
 16 #define y second
 17 #define clr(a,b) memset(a,b,sizeof(a))
 18 #define For(i,n) for(int i=1;i<=n;i++)
 19 #define Rep(i,l,r) for(int i=l;i<=r;i++)
 20 #define Swap(A,B) {Points T = A ; A = T; T = B;}
 21 int n,npcnt,es,stack[5*N],head[N*5],dis[5*N],ans;
 22 set<Points> hash,farm;
 23 bool vis[5*N];
 24 Points p[N],np[5*N];
 25 struct edge{
 26     int t,next,w;
 27 }E[5*N*5*N];
 28 
 29 int Len(Points A,Points B){
 30     return abs(B.x-A.x)+abs(B.y-A.y);
 31 }
 32 
 33 void makelist(int s,int t){
 34     E[es].t = t; E[es].next = head[s];E[es].w = Len(np[s],np[t]);
 35     head[s] = es++;
 36 }
 37 
 38 bool inseg(Points out,Points A,Points B){
 39     if(A.x==B.x)
 40             if(min(A.y,B.y)<out.y&&out.y<max(A.y,B.y)&&out.x==A.x) return true;
 41     if(A.y==B.y)
 42             if(min(A.x,B.x)<out.x&&out.x<max(A.x,B.x)&&out.y==A.y) return true;
 43     return false;
 44 }
 45 
 46 bool check(Points A,Points B){
 47     Points C;//   _| 
 48     if(A.x>B.x) Swap(A,B);
 49     C.y = A.y; C.x = B.x;
 50     bool Flag = A.x==B.x||A.y==B.y||farm.find(C)==farm.end();
 51     For(i,n)
 52         if(inseg(p[i],A,C)||inseg(p[i],C,B)){
 53             Flag = false;
 54             break;
 55         }
 56     if(Flag) return true;
 57     C.x = A.x; C.y = B.y;
 58     Flag = A.x==B.x||A.y==B.y||farm.find(C)==farm.end();
 59     For(i,n)
 60         if(inseg(p[i],A,C)||inseg(p[i],C,B)){
 61             Flag = false;
 62             break;
 63         }
 64     return Flag;
 65 }
 66 
 67 void init(){
 68     scanf("%d",&n);
 69     For(i,n) {
 70         scanf("%d%d",&p[i].y,&p[i].x);
 71         hash.insert(p[i]);farm.insert(p[i]);
 72         np[i] = p[i];
 73     }npcnt = n;
 74     clr(head,-1);
 75     For(i,n)
 76         For(j,4){
 77             int newx = p[i].x + dx[j] , newy = p[i].y + dy[j];
 78             if(newx<1||newx>limits||newy<1||newy>limits) continue;
 79             if(hash.find(make_pair(newx,newy))==hash.end()) {np[++npcnt] = make_pair(newx,newy);hash.insert(make_pair(newx,newy));}
 80         }
 81     For(i,npcnt-1)
 82       Rep(j,i+1,npcnt){
 83             if(check(np[i],np[j])) {
 84                 makelist(i,j);
 85                 makelist(j,i);
 86             }
 87       } 
 88 }
 89 
 90 int SPFA(int s,int t){
 91     For(i,5*n) dis[i] = 214748364;
 92     memset(vis,0,sizeof(vis));
 93     dis[s] = 0;
 94     stack[stack[0]=1,1] = s;
 95     while(stack[0]){
 96         int First = stack[stack[0]--];vis[First] = false;
 97         for(int p=head[First];p!=-1;p=E[p].next){
 98             int v = E[p].t;
 99             if(v<=n&&v!=s&&v!=t) continue;
100             if(dis[First]+E[p].w<dis[v]) {
101                 dis[v] = dis[First] + E[p].w;
102                 if(!vis[v]) {vis[v] = true; stack[++stack[0]] = v;}
103             } 
104         }  
105     }
106     return dis[t];
107 }
108 
109 void Work(){
110     For(i,n){
111         int W = SPFA(i,(i==n)?(1):(i+1)); 
112         if(W>21474836) puts("-1");
113         if(W>21474836) return;
114         ans+=W;
115     }
116     printf("%d\n",ans);
117 } 
118 
119 int main(){
120    // freopen("date.in","r",stdin);
121    // freopen("date.out","w",stdout);
122     init();
123     Work();
124     return 0;
125 }
126 --------------------------------------A题--------------------------------------
127 #include<queue>
128 #include<cstdio>
129 #include<cstdlib>
130 #include<cstring>
131 #include<iostream>
132 #include<algorithm>
133 using namespace std;
134 const int N = 30;
135 const int MAXS = 20*100/3;
136 #define For(i,n) for(int i=1;i<=n;++i)
137 #define Rep(i,l,r) for(int i=l;i<=r;++i)
138 
139 int A[N],f[N][MAXS+200][MAXS+200],sum,ans=2147483647,n;
140 
141 int main(){
142     scanf("%d",&n);
143     For(i,n) {scanf("%d",&A[i]),sum+=A[i];};
144     memset(f,false,sizeof(f));
145     f[0][0][0] = true;
146     For(i,n){
147         Rep(j,0,MAXS)
148           Rep(k,0,MAXS) 
149             if(f[i-1][j][k]){
150                 f[i][j][k]      = true;
151                 f[i][j+A[i]][k] = true;
152                 f[i][j][k+A[i]] = true;
153             }
154     }
155     For(i,MAXS){
156       For(j,MAXS)
157         if(f[n][i][j])
158             ans = min(ans,max(i,max(j,sum-i-j)));
159     }
160     printf("%d\n",ans);
161     return 0;
162 }
163 --------------------------------------B题--------------------------------------
164 #include<queue>
165 #include<cstdio>
166 #include<iostream>
167 #include<algorithm>
168 using namespace std;
169 const int N = 25100;
170 #define Points pair<int,int>
171 #define For(i,n) for(int i=1;i<=n;++i)
172 #define Rep(i,l,r) for(int i=l;i<=r;++i)
173 
174 Points A[N];
175 int n;
176 
177 bool cmp(Points A,Points B){
178     if(A.first<A.second){
179         if(B.first<B.second) return A.first<B.first;
180         else                 return true;
181     }else{
182         if(B.first>B.second) return A.second>B.second;
183         else                 return false;
184     }    
185 }
186 
187 void solve(){
188     sort(A+1,A+n+1,cmp);
189     int sum1 = 0 ,sum2 = A[1].first;
190     For(i,n){
191         sum1+=A[i].first;
192         sum2=max(sum1,sum2)+A[i].second;
193     }
194     printf("%d\n",sum2);
195 }
196 
197 int main(){
198     scanf("%d",&n);
199     For(i,n) scanf("%d%d",&A[i].first,&A[i].second);
200     solve();
201     return 0;
202 }
203 ----------------------------------C题------------------------------------------